Download QUESTION Solve the following differential equations subject to the

QUESTION
Solve the following differential equations subject to the given initial conditions.
dy
= x + 6, where y = 3 when x = 0.
(i) (x + 1) dx
dy
(ii) x dx
+ (3x + 1)y = e−3x , where y = e−3 when x = 1.
2
d y
dy
(iii) 2 dx
2 − 5 dx − 3y = 0, where y = 1 and
(iv)
d2 y
dx2
dy
dx
dy
− 10 dx
+ 25y = 0, where y = 1 and
= 3 when x = 0.
dy
dx
= 1 when x = 1.
ANSWER
(i)
dy
x+6
5
=
=1+
dx
x+1
x+1
⇒ y = x + 5 ln |x + 1| + c 3 = 5 ln 1 + c so c = 3
⇒ y = x + 5 ln |x + 1| + 3
(ii)
dy
dx
³
+ 3+
1
x
´
y=
1
xe3x
Integrating factor: e
R
3+ x1 dx
= e(3x+ln |x|) = |x|e3x . So for x > 0,
d(xe3x y)
1
= 3x .xe3x ⇒ xe3x y
dx
xe
Z
so y =
1 dx ⇒ y =
x+C
c+1
: e−3 = 3 ⇒ C = 0
3x
xe
e
1
e3x
(iii) Auxiliary equation: 2λ2 − 5λ − 3 = 0 factorises as (2λ + 1)(λ − 3) so
1
λ = 3, − 12 with solutions y = Ae3x + Be− 2 x . 1 = A + B, 3 = 3A − 12 B
from initial conditions so 3 12 B = 0 ⇒ B = 0 and A = 1 giving y = e3x .
(iv) Auxiliary equation: λ2 − 10λ + 25 = 0 factorises as λ − 5)2 so there is
a unique solution λ = 5. y = (A + Bx)e5x .
dy
= Be5x + 5(A + Bx)e5x = (5A + B + 5Bx)e5x
dx
1 = (A + B)e5 and 1 = (5A + 6B)e5 from the initial conditions so
5 = Ae5 or A = 5e−5 and so Be5 = 1 − 5e−5 i.e B = −4e−5 and
y = (5 − 4x)e5(x−1)
1